Simplex Repulsion
Hackett, C.A. et al. 1998. Linkage analysis in tetraploid species:
a simulation study. Genet. Res. Comb. 71:143-154.
The four homologous chromosomes have the following genotypes:
Xy = A
xY = B
xy = C
xy = D
When A pairs with C or D, there is no observed recombination. When
B pairs with C or D, there is no observed recombination. Only when A
and B pair, would observed recombination occur.
| X y |
|
xy |
|
| o |
|
o |
Case
1 |
| |
|
|
A+B |
| o |
|
o |
C+D |
| x Y |
|
xy |
|
| X y |
|
xY |
|
| o |
|
o |
Case
2 |
| |
|
|
A
+ C |
| o |
|
o |
B
+ D |
| x y |
|
xy |
|
| X y |
|
xY |
|
| o |
|
o |
Case
3 |
| |
|
|
A
+ D |
| o |
|
o |
B
+ C |
| x y |
|
xy |
|
Of the six possible bivalent pairs, only when A pairs with B would
recombination be observable. When A and B pair, the combinations produced
would be:
| ( |
1-r |
Xy. |
1-r |
xY: |
r |
XY: |
r |
xy |
) |
| 2 |
2 |
2 |
2 |
When C pairs with D, the combinations produced
would be: (xy)
For Case 1 we would have the following gametes:
| 1 - r |
Xyxy. |
1 - r |
xYxy |
| 2 |
2 |
r/2XYxy:r/2xyxy
When A pairs with C, the combinations would be:
| 1-r |
Xy. |
1-r |
xy. |
r |
Xy. |
r |
xy |
| 2 |
2 |
2 |
2 |
which reduces to:
| ( |
1-r |
+ |
r |
) |
Xy+ |
( |
1-r |
+ |
r |
) |
xy= |
1 |
Xy+ |
1 |
xy. |
| 2 |
2 |
2 |
2 |
2 |
2 |
When B pairs with D, the combinations would be:
| 1-r |
xY: |
1-r |
xy. |
r |
xY: |
r |
xy |
| 2 |
2 |
2 |
2 |
which reduces to ½ xY: ½ xy.
For Case 2 we would have the following gametes:
¼ XyxY: ¼ Xyxy: ¼xyxY: ¼ xyxy.
For Case 3 we would have the following gametes:
¼ XyxY: ¼ Xyxy: ¼ xyxY: ¼ xyxy:
| To
summarize |
| Case |
Gametes |
| 1 |
1-r |
Xxyy. |
1-r |
xxYy. |
r |
xxyy |
| 2 |
2 |
2 |
| 2 |
¼
XxYy: ¼ Xxyy: ¼ xxYy: ¼ xxyy |
| 3 |
¼
XxYy: ¼ Xxyy: ¼ xxYy: ¼ xxyy |
| Gametes |
Probability |
Observed
Number |
| Xxyy |
[ |
1-r |
+ |
1 |
+ |
1 |
] |
1 |
=(2-r)/6 |
b |
| 2 |
4 |
4 |
3 |
| xxYy |
[ |
1-r |
+ |
1 |
+ |
1 |
] |
1 |
=(2-r)/6 |
c |
| 2 |
4 |
4 |
3 |
| XxYy |
[ |
r |
+ |
1 |
+ |
1 |
] |
1 |
= |
(1+r) |
|
a |
| 2 |
4 |
4 |
3 |
6 |
|
| xxyy |
[ |
r |
+ |
1 |
+ |
1 |
] |
1 |
= |
(1+6) |
|
d |
| 2 |
4 |
4 |
3 |
6 |
The probability of observing a, b, c, and d of each class is:
| P |
( |
a,b,c,d; |
(1+r) |
, |
(2-r) |
, |
(2-r) |
, |
(1+r) |
) |
| 6 |
|
6 |
|
6 |
|
6 |
| = |
n! |
( |
1+r |
) |
a |
( |
2-r |
) |
b |
( |
2-r |
) |
c |
( |
1+r |
) |
d |
| a!b!c!d! |
6 |
|
6 |
|
6 |
|
6 |
|
We can solve for r by taking the derivative of the log of this function
with respect to r and setting the derivative equal to zero. The log-likelihood,
L, is given by:
| L(r)=log |
[ |
n! |
] |
+a log |
[ |
1+r |
] |
+b log |
[ |
2-r |
] |
+c log |
[ |
2-r |
] |
+d |
[ |
1+r |
] |
| a!b!c!d! |
6 |
6 |
6 |
6 |
| d[L(r)] |
=0+ |
a |
- |
b |
- |
c |
+ |
d |
|
|
| dr |
(1+r)/6 |
(2-r)/6 |
(2-r)/6 |
(1+r)/6 |
|
|
| a+d |
= |
b+c |
|
|
|
|
|
|
|
|
| (l+r)/6 |
(2-r)/6 |
|
|
|
|
|
|
|
|
| (a+d)(2-r) |
= |
(b+c)(1+r) |
|
|
|
|
|
|
|
|
| 6 |
6 |
|
|
|
|
|
|
|
|
2a + 2d - ra- rd = b + c + br + cr
2a - b - c + 2d = r(a+b+c+d)
2a - b - c + 2d = rn, because n = a + b + c + d
| 2a-b-c+2d |
= r; |
which is the solution |
| n |
Copyright
2000©, Ted Helms | 
